In a single throw of two dice, find the probability of
Question:

In a single throw of two dice, find the probability of

(i) getting a sum less than 6

(ii) getting a doublet of odd numbers

(iii) getting the sum as a prime number

 

Solution:

(i) We know that,

Probability of occurrence of an event

$=\frac{\text { Total no. of Desired outcomes }}{\text { Total no.of outcomes }}$

Outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Total no.of outcomes are 36

In that only $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)$ are our desired outputs as there sum is less than 6

Therefore no.of desired outcomes are 10

Therefore, the probability of getting a sum less than 6

$=\frac{10}{36}$

$=\frac{5}{18}$

Conclusion: Probability of getting a sum less than 6, when two dice are rolled is $\frac{5}{18}$

(ii) We know that,

Probability of occurrence of an event

$=\frac{\text { Total no. of Desired outcomes }}{\text { Total no. of outcomes }}$

In $(a, b)$ if $a=b$ then it is called a doublet

Total doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

In $(a, b)$ if $a=b$ and if $a, b$ both are odd then it is called a doublet

Odd doublets are $(1,1),(3,3),(5,5)$

Outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Total no. of outcomes are 36 and desired outcomes are 3

Therefore, probability of getting doublet of odd numbers

$=\frac{3}{36}$

$=\frac{1}{12}$

Conclusion: Probability of getting doublet of odd numbers, when two dice are rolled

is $\frac{1}{12}$

(iii) We know that,

Probability of occurrence of an event

$=\frac{\text { Total no.of Desired outcomes }}{\text { Total no. of outcomes }}$

Outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Total no. of outcomes are 36

Desired outputs are $(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1)$, $(4,3),(5,2),(5,6),(6,1),(6,5)$

Total no. of desired outputs are 15

Therefore, probability of getting the sum as a prime number

$=\frac{15}{36}$

$=\frac{5}{12}$

Conclusion: Probability of getting the sum as a prime number, when two dice are rolled

is $\frac{5}{12}$

 

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